# Homework Help: Do I need to know the height to solve this?

1. Jun 26, 2011

### flyingpig

1. The problem statement, all variables and given/known data

When a 5kg mass is hung vertically on a light spring the spring stretches 3 cm. how much work must a external agent do to stretch the spring 4 cm from its equilibrium position?

Because I am guessing it is hung vertically from the ceiling?

2. Jun 26, 2011

### fluidistic

No you don't need the height.
Yes it's hung vertically from the ceiling. I'll give you a hint to solve the problem: consider the potential energy stored in a string.

3. Jun 26, 2011

### flyingpig

Why don't you need to know the height? Isn't there gravity? I know that it is just

$$\int_{x_0 = 3cm}^{x = 4cm} -kx dx$$

But I am sure that I need gravity

4. Jun 26, 2011

### Fewmet

You need the acceleration due to gravity, with varies do little with height that it is treated as a constant 9.81 m/s2 near the surface of the Earth.

Do you see how to calculate k from the given information?

Also note that the question asks for the work done stretching from the equilibrium position to 4 cm. The integral you posted is inconsistent with that.

5. Jun 26, 2011

### fluidistic

This is right.
Although I do find the wording of the question confusing. When they say
, do they mean equilibrium position when there's the mass attached to the spring or when the mass isn't attached yet? Because if it is the former then flyingpig is right, I believe.

6. Jun 27, 2011

### flyingpig

[PLAIN]http://img824.imageshack.us/img824/403/hieh.jpg [Broken]

I need to know height, this is a picture of how I interpret the problem

Last edited by a moderator: May 5, 2017
7. Jun 27, 2011

### Sourabh N

Let me rephrase the question. This might make it clear what the question is asking.

How much work must a external agent do to stretch a light spring 4 cm from its equilibrium position? Oh by the way, when a 5kg mass is hung vertically on the spring the spring stretches 3 cm.

8. Jun 27, 2011

### Fewmet

That's how I was seeing it. The "by the way" clause let's you find k, and that then gets you an answer.

9. Jun 27, 2011

### fluidistic

Ohhh... I see now. Personally I thought flyingpig had its limits of the integral right. So thanks a lot for the clarification.
And who else agree with me when I say that flyingpig doesn't need the height? As long as g is treated as a constant, there's no need to know the height.

10. Jun 28, 2011

### flyingpig

What...?

11. Jun 28, 2011

### Sourabh N

Read the posts after yours and point out the line which you do not understand.

12. Jun 28, 2011

### flyingpig

Isn't that what I did with my integral? BUt my integral is still only half completed.

13. Jun 28, 2011

### Fewmet

I see I was mistaken about the integral. At a glance I thought you were using it to find the work, but now I see that expression is intended to equal force. IN that case the limits of integration are correct.

14. Jun 28, 2011

### PieOperator

Although springs exert a nonconservative force, you can treat this problem as a conservative force problem for now. Conservative forces have a definite integral and a potential energy graph. Thus, you need to know the height when using the ever easy equation of potential energy equals mass times gravity times height. Just think about how you can apply the 2nd law of thermodynamics into this great and insightful setting!

15. Jun 28, 2011

### fluidistic

Look at post #6. When you say "you need to know the height when using the ever easy equation of potential energy equals mass times gravity times height.", flyingpig might think he needs to know the value of d or h. And he doesn't need these values.

16. Jun 29, 2011

### flyingpig

Okkay, how about this, tell me why don't you need the height? Why is there no gravity in play? It's hanging vertically

17. Jun 29, 2011

### Fewmet

I suspect you are thinking about this differently than some of us, and the confusion is coming from that. Can you say how you would use the height if it were given? That should let us explain why it is not necessary.

18. Jun 30, 2011

### flyingpig

[PLAIN]http://img200.imageshack.us/img200/92/unledvx.jpg [Broken]

$$\sum W = \int_{x_0 = 3\cdot 10^{-2}m}^{x = 4\cdot 10^{-2}m} -kx dx + -mg(5m - 6m)$$

Last edited by a moderator: May 5, 2017
19. Jun 30, 2011

### Fewmet

Here is the original question:
The work done by the "outside agent" depends on the force he/she/it applies to the 5 kg mass. The -kx is the upward force exerted by the spring. The mg is the force exerted by gravity. Where is the force exerted by the outside agent?

20. Jun 30, 2011

### fluidistic

I'm afraid this sketch is wrong. Seems like you mean that the weight of the mass stretched the spring 1 meter while this isn't true.
Also we can see that 1cm=1m from the sketch.

So now we interpret the problem as stretching the spring 4 cm from its equilibrium position (when there's no mass attached) when a mass of 5 kg is hanging from the spring?

Last edited by a moderator: May 5, 2017
21. Jun 30, 2011

### Fewmet

That's how I am reading it, fluidistic, although I wonder if flyingpig rephrased a little when transcribing it to the original post.

flyingpig: Something further occurred to me as I thought more about this. You supposed the heights were 5 m and 6 m (although fluidistic is right that the stretch is only 1 cm). Do you see that no matter what heights you choose the ∆x will still be 1 cm? That's why you don;t need to know the height, just the change in height.

22. Jun 30, 2011

### fluidistic

Thanks Fewmet.
Now hopefully flyinpig will understand why he doesn't need the height.

23. Jul 2, 2011

### flyingpig

OKay sorry I meant cm,

So let's just pretend the picture had 6m = 6cm and 5m = 5cm.

Fewmet said we only need the change in height, but fluid said we don't need it at all??

24. Jul 2, 2011

### Fewmet

You don't need the height in that whether this was 1 m off the follow or 50 m, the answer will still be the same. You just need that the mass is getting 1 cm farther from the equilibrium position.

So you want to know how much work the "outside agent" does in pulling the mass that extra 1 cm. That means you need an expression for the the force applied that did that work.

You are really in the verge of getting an answer.

I will be away from the computer for the next 6 hours or so, so I'll not be posting replies anytime soon.

25. Jul 2, 2011

### omega_minus

Flyingpig, have you tried a free body diagram yet?